Advertisements
Advertisements
प्रश्न
Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.
Advertisements
उत्तर
L.H.S. = (1 – cos2A) . sec2B + tan2B (1 – sin2A)
= `sin^2A * 1/(cos^2B) + (sin^2B)/(cos^2B) (1 - sin^2A)` ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= `(sin^2A)/(cos^2B) + (sin^2B)/(cos^2B) - (sin^2A sin^2B)/(cos^2B)`
= `(sin^2A)/(cos^2B) - (sin^2A sin^2B)/(cos^2B) + (sin^2B)/(cos^2B)`
= `(sin^2A)/(cos^2B) (1 - sin^2B) + tan^2B`
= `(sin^2A)/(cos^2B) (cos^2B) + tan^2B`
= sin2A + tan2B
= R.H.S.
∴ (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Find the value of sin 30° + cos 60°.
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
sec 60° = ?
If `sec θ = 41/40`, then find values of sin θ, cot θ, cosec θ.
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
